Which of the following is a galvanic cell?
2 answers:
C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.
or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
<em />
From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
You might be interested in
Explanation:
According to the Henderson-Hasselbalch equation, the relation between pH and is as follows.
pH =
where, pH = 7.4 and = 7.21
As here, we can use the nearest to the desired pH.
So, 7.4 = 7.21 +
0.19 =
= 1.55
1 mM phosphate buffer means +
= 1 mM
Therefore, the two equations will be as follows.
= 1.55 ............. (1)
+
= 1 mM ........... (2)
Now, putting the value of from equation (1) into equation (2) as follows.
1.55 = 1 mM
2.55 = 1 mM
= 0.392 mM
Putting the value of in equation (1) we get the following.
0.392 mM + = 1 mM
= (1 - 0.392) mM
= 0.608 mM
Thus, we can conclude that concentration of the acid must be 0.608 mM.